Question: A bag contains $2$ red marbles, $3$ green marbles, and $4$ blue marbles. If we choose a marble, then another marble without putting the first one back in the bag, what is the probability that the first marble will be green and the second will be red?
Solution: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is picking a green marble and leaving it out. Event B is picking a red marble. Let's take the events one at at time. What is the probability that the first marble chosen will be green? There are $3$ green marbles, and $9$ total, so the probability we will pick a green marble is $\dfrac{3} {9}$. After we take out the first marble, we don't put it back in, so there are only $8$ marbles left. Since the first marble was green, there are still $2$ red marbles left. So, the probability of picking a red marble after taking out a green marble is $\dfrac{2} {8}$. Therefore, the probability of picking a green marble, then a red marble is $\dfrac{3}{9} \cdot \dfrac{2}{8} = \dfrac{6}{72} = \dfrac{1}{12}$